1304 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Since ε > 0 is arbitrary, it follows u = 0 a.e. Next suppose ui ∈ Hm (U) for i = 1,2. Thereexists vi ∈ Hm (Rn) such that

||vi||Hm(Rn) < ||ui||Hm(U)+ ε.

Therefore,

||u1 +u2||Hm(U) ≤ ||v1 + v2||Hm(Rn) ≤ ||v1||Hm(Rn)+ ||v2||Hm(Rn)

≤ ||u1||Hm(U)+ ||u2||Hm(U)+2ε

and since ε > 0 is arbitrary, this shows the triangle inequality.The interesting question is the one about completeness. Suppose then {uk} is a Cauchy

sequence in Hm (U) . There exists Nk such that if k, l ≥ Nk, it follows ||uk−ul ||Hm(U) <12k

and the numbers, Nk can be taken to be strictly increasing in k. Thus for

l ≥ Nk,∣∣∣∣ul−uNk

∣∣∣∣Hm(U)

< 1/2l .

Therefore, there exists wl ∈ Hm (Rn) such that

wl |U = ul−uNk , ||wl ||Hm(Rn) <12l .

Also let vNk |U = uNk with vNk ∈ Hm (Rn) and

∣∣∣∣vNk

∣∣∣∣Hm(Rn)

<∣∣∣∣uNk

∣∣∣∣Hm(U)

+12k .

Now for l > Nk, define vl by vl−vNk = wNk so that∣∣∣∣vl− vNk

∣∣∣∣Hm(Rn)

< 1/2k. In particular,∣∣∣∣vNk+1 − vNk

∣∣∣∣Hm(Rn)

< 1/2k

which shows that{

vNk

}∞

k=1 is a Cauchy sequence. Consequently it must converge to v ∈Hm (Rn) . Let u = v|U . Then∣∣∣∣u−uNk

∣∣∣∣Hm(U)

≤∣∣∣∣v− vNk

∣∣∣∣Hm(Rn)

which shows the subsequence,{

uNk

}k converges to u. Since {uk} is a Cauchy sequence, it

follows it too must converge to u. This proves the lemma.The main result is next.

Theorem 38.1.8 Suppose U satisfies Assumption 38.1.2. Then for m a nonnegative integer,Hm (U) =W m,2 (U) and the two norms are equivalent.

Proof: Let u ∈ Hm (U) . Then there exists v ∈ Hm (Rn) such that v|U = u. Hencev ∈W k,2 (Rn) and so all its weak derivatives up to order m are in L2 (Rn) . Therefore, therestrictions of these weak derivitves are in L2 (U) . Since U satisfies the segment condition,it follows u ∈W m,2 (U) which shows Hm (U)⊆W m,2 (U) .